Physics · Electric Charges And Fields · NEET
A thin sheet radiates charge from BOTH of its faces, so the field splits equally into two sides. In the Gauss derivation the flux comes out of two flat caps (2EA), giving 2EA = σA/ε₀, so E = σ/2ε₀. You only get σ/ε₀ for a thick CONDUCTOR surface, where charge sits on one outer face and there is no field inside the metal, so all the flux goes out one side.
For an infinite sheet the field lines are parallel and never spread out, so the field never weakens with distance. E = σ/2ε₀ is a constant (uniform field). This is unlike a point charge (E ∝ 1/r²) or a line charge (E ∝ 1/r). NEET loves this: moving a test charge farther from a big sheet does NOT change the field.
Take +σ and −σ sheets. Between them the two fields point the same way and add: E = σ/2ε₀ + σ/2ε₀ = σ/ε₀. Outside the pair the two fields cancel, so E = 0 outside. This is exactly the parallel-plate capacitor result — remember: INSIDE = σ/ε₀, OUTSIDE = 0.
Straight out, perpendicular (normal) to the sheet. For a positive sheet the field points away from the sheet on both sides; for a negative sheet it points toward the sheet. By symmetry it can only be normal to the plane — any sideways component would cancel.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
E = σ/2ε₀, where σ is the surface charge density (C/m²) and ε₀ is the permittivity of free space. It is directed normal to the sheet, away from a positive sheet.
No. E = σ/2ε₀ is uniform — it is the same at every distance from the sheet. This is why an infinite sheet is treated as a source of a uniform electric field.
A cylinder (or rectangular box) with its flat faces parallel to the sheet, piercing it symmetrically. Only the two flat caps carry flux (2EA); the curved side is parallel to the field and carries none.
E = σ/ε₀ between the plates (fields add) and zero outside. This is the parallel-plate capacitor field.
In a conductor the field inside is zero and charge lives on one outer face, so all flux exits one cap → σ/ε₀. A thin sheet pushes field out of both faces → σ/2ε₀.
